# Changes to multivariate AR1 parameters with step size

Given a univariate AR1 model with step size $$n$$

$$X(t+n) = a_n + b_n X(t) + \sigma_n Z(t+n),$$

we can change to a model with a different step size $$m$$

$$X(t+m) = a_m + b_m X(t) + \sigma_m Z(t+m),$$

using

$$b_m = b_n^{m/n}$$ $$a_m = a_n \frac{1 - b_m}{1 - b_n}$$ $$\sigma_m = \sigma_n \sqrt\frac{1 - b_m^2}{1 - b_n^2}$$

But what if we have a multivariate model with, say, two AR1 processes $$X(t)$$ and $$Y(t)$$ where the innovations have a correlation of

$$Corr[Z_X(t+n),Z_Y(t+n)] = \rho_n$$

Is this correlation parameter affected by the step size change? How can we get from $$\rho_n$$ to $$\rho_m$$?

Add 1

The relationship between parameters for the univariate model are obtained by equating expressions for the unconditional mean, variance, and auto-correlation, i.e.

$$E[X(t)] = \frac{a_n}{1 - b_n} = \frac{a_m}{1 - b_m}$$ $$Var[X(t)] = \frac{\sigma_n^2}{1 - b_n^2} = \frac{\sigma_m^2}{1 - b_m^2}$$ $$Cov[X(t), X(t - m \times n)] = b_n^m Var[X(t)] = b_m^n Var[X(t)]$$

Add 2

I suppose, similarly to the uni-variate derivation above, one can equate the expressions for cross-covariance between the two processes, i.e.

$$Cov[X(t),Y(t)] = \frac{\rho_n \sigma_{x;n} \sigma_{y;n} }{1 - b_{x;n} b_{y;n}} = \frac{\rho_m \sigma_{x;m} \sigma_{y;m} }{1 - b_{x;m} b_{y;m}}$$

which would imply that

$$\rho_m = \rho_n \frac{\sigma_{x;n} \sigma_{y;n}}{\sigma_{x;m} \sigma_{y;m}} \frac{1 - b_{x;m} b_{y;m}}{1 - b_{x;n} b_{y;n}} = \rho_n \frac{1 - b_{x;m} b_{y;m}}{1 - b_{x;n} b_{y;n}} \sqrt\frac{\left(1 - b_{x;n}^2\right)\left(1 - b_{y;n}^2\right)}{\left(1 - b_{x;m}^2\right)\left(1 - b_{y;m}^2\right)}$$

but I am not sure if this is the right way to do it.

• Hi: could you explain or show the derivation of the $m$ parameters ? If you do that, someone might know how to do the correlation. Although someone might also know without that being shown :). Commented Aug 25, 2020 at 12:27
• It's not clear to me how you obtained the expression for the cross correlation of X(t) and Y(t) but, assuming it's correct, I would think that that way would be the way to go. You seem to have answered your own question :). Commented Aug 26, 2020 at 12:30